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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

K _ 1 and K _ 2 are the respective equilibrium constant for the...

K1{K}_{1} and K2{K}_{2} are the respective equilibrium constant for the reactions
(i) XeF6(g)+H2O(g)⇌XeOF4(g)+2HF(g);K1Xe{F}_{6} ( g ) + {H}_{2}O ( g ) ⇌ XeO{F}_{4} ( g ) + 2HF ( g ) ; {K}_{1}
(ii) XeO4(g)+XeF6(g)⇌XeOF4(g)+XeO3F2;K2Xe{O}_{4} ( g ) + Xe{F}_{6} ( g ) ⇌ XeO{F}_{4} ( g ) + Xe{O}_{3}{F}_{2}; {K}_{2}
The equilibrium constant for the reaction
XeO4(g)+2HF(g)⇌XeO3F2(g)+H2O(g)Xe{O}_{4} ( g ) + 2HF ( g ) ⇌ Xe{O}_{3}{F}_{2} ( g ) + {H}_{2}O ( g ) will be

Options

  1. AK1K22\dfrac{{K}_{1}}{{K}_{2}^{2}}
  2. BK1−K2{K}_{1} - {K}_{2}
  3. CK1K2\dfrac{{K}_{1}}{{K}_{2}}
  4. DK2K1\dfrac{{K}_{2}}{{K}_{1}}

Correct answer

D. K2K1\dfrac{{K}_{2}}{{K}_{1}}

Step-by-step solution

(i) XeF6(g)+H2O(g)⇌XeOF4(g)+2HF(g);K1Xe{F}_{6} ( g ) + {H}_{2}O ( g ) ⇌ XeO{F}_{4} ( g ) + 2HF ( g ) ; {K}_{1}
(ii) XeO4(g)+XeF6(g)⇌XeOF4(g)+XeO3F2;K2Xe{O}_{4} ( g ) + Xe{F}_{6} ( g ) ⇌ XeO{F}_{4} ( g ) + Xe{O}_{3}{F}_{2}; {K}_{2}
Required equation
XeO4(g)+2HF(g)⇌XeO3F2(g)+H2O(g)Xe{O}_{4} ( g ) + 2HF ( g ) ⇌ Xe{O}_{3}{F}_{2} ( g ) + {H}_{2}O ( g )
It is obtained by subtracting (i) from (ii), hence equilibrium constant of the required equation will be
K2K1\dfrac{{K}_{2}}{{K}_{1}}.
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